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\begin{document}

\centerline{\huge
\textbf{Complex Map} $z \mapsto {1\over z}$}
\bigskip
\centerline{(\textit{From the 3D-XplorMath Project, \url{http://3d-xplormath.org}})}

%  Section headings and comments below are suggestions for further discussion...
%\heading{About This Exhibit}


%  BASIC DESCRIPTION OF THE MATHEMATICAL OBJECT GOES HERE.
%  (Eg. What is the object?  What equations defines it?  Why is it interesting?)

\Large{
See the discussion in ``About this Category'' for details on what to look at, what to expect, and what to do.
We recall that when an example is selected in the Conformal Map Category, first a grid is shown
and then the image of that grid under the conformal map.
The first conformal map examples to look at, (using Cartesian  {\bf and} Polar
Grids) are $ z\to z^2,\ z\to 1/z,\ z\to \sqrt{z},\ z\to e^z$.
}

%  Discussion of what is shown on the screen, parameters, default morphs.
\Large{
REMARK: The actual mapping for this example is 
$z \mapsto {aa \over(z - bb)} + cc$, 
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$aa = 1$, $bb = 0$, $cc = 0$. The default morph goes from $bb = 0$ to $bb = 1$.
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\LARGE
Look at the function $z\mapsto z^2$ and its ATO first.

The function $ z \mapsto 1/z$ should be looked at  both in Cartesian
and Polar Grids.


Notice first: 
\begin{list}{}{}
\item {1)} The real axis, imaginary axis and unit circle are mapped
into themselves,   

\item {2)} the upper half plane and the lower half plane
are interchanged, and 

\item {3)} the inside of the unit circle and its outside
are also interchanged. 
\end{list}

This is best seen in the (default) Conformal Polar Grid.
In the Cartesian Grid one should in particular observe that all
straight parameter lines (in the domain) are mapped to circles
(some exceptions, like the real axis, remain lines). The behaviour
of these circles near zero can be looked at as an image of the
behaviour of the standard Cartesian Grid near infinity. In fact {\bf
all} circles are mapped to circles or lines.

Examples to look at after this are 

\noindent
$\ \ z\mapsto (az+b)/(cz+d)$ and $z\mapsto (z+cc)/(1+\bar{cc}z)$,

both of which can be obtained from $z\to 1/z$ by composition with
translations $z\to z+a$ or scaled rotations $z\to a\cdot z$. 
Therefore all of these so-called 
``M\"obius transformations'' map circles and lines to circles
and lines. 

\author{H.K.}


%\heading{About This Exhibit}
%  Discussion of what is shown on the screen, parameters, default morphs.

%\heading{Things to Try}
%  Suggestions for other things that can be tried with the exhbit.

%\heading{More About the Math}
%  More advanced mathematical treatment and other materials could go here.

%\heading{More Information}
%\Large
%  Links to material on the Internet could go here
%  Example:  See \href{http://3d-xplormath.org}{the 3D-XplorMath web site}

\end{document}


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